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1
8
y∂
x
ϕ
1
2
β
2
2
2
b
1
(ϕ, ϕ)
=
L
2
(G)
,
2
(ϕ, ϕ)
=
∂
y
ϕ
L
2
(G)
,
1
4
[
2
β
2
κ
2
b
6
(ϕ, ϕ)
=
−
α
]
ϕ
L
2
(G)
,
1
2
κ
κβ
2
2
2
b
8
(ϕ, ϕ)
=
[
α
−
]
yϕ
L
2
(G)
,
9
(ϕ, ϕ)
=[
r
−
2
καm
]
ϕ
L
2
(G)
,
1
/
2
(∂
x
(ϕ)
2
,
1
)
1
/
2
(∂
x
(ϕ)
2
,y
2
)
and, using
(∂
x
ϕ,ϕ)
=
=
0,
(y∂
x
ϕ,yϕ)
=
=
0, we
find that
b
4
(ϕ, ϕ)
=
b
5
(ϕ, ϕ)
=
0. Furthermore, for
ε>
0,
1
2
|
b
3
(ϕ, ϕ)
≥−
ρβ
|
y∂
x
ϕ
L
2
(G)
∂
y
ϕ
L
2
(G)
ρ
2
β
2
4
ε
1
2
ε
1
2
2
2
≥−
y∂
x
ϕ
L
2
(G)
−
∂
y
ϕ
L
2
(G)
,
1
2
|
(
9.25
)
≥−|
β
2
y
−
1
ϕ
β
2
2
b
7
(ϕ, ϕ)
≥−
4
αm
−
|
∂
y
ϕ
L
2
(G)
L
2
(G)
4
αm
−
|
∂
y
ϕ
L
2
(G)
.
Collecting these estimates yields
a
κ
(ϕ, ϕ)
2
2
2
≥
c
1
y∂
x
ϕ
L
2
(G)
+
c
2
∂
y
ϕ
L
2
(G)
+
c
3
yϕ
L
2
(G)
+
c
4
ϕ
L
2
(G)
,
β
2
ρ
2
4
ε
1
1
1
1
2
1
2
β
2
β
2
κβ
2
with
c
1
=
8
−
2
ε
,
c
2
=
−
−|
4
αm
−
|
,
c
3
=
2
κ
[
α
−
]
and
c
4
=
1
1
4
2
β
2
κ
r
−
2
καm
+
4
[
−
α
]
. We choose
ε
sufficiently close to
such that
c
1
>
0.
Thus, in order for
c
2
to be positive, we must have
1
2
β
2
>
1
2
β
2
ρ
2
β
2
1
>ρ
2
4
αm/β
2
+|
4
αm
−
|⇔
+
2
|
−
1
|
.
Furthermore, if 0
<κ<α/β
2
, then we have
c
3
>
0. Hence,
1
a
κ
(ϕ, ϕ)
2
2
L
2
(G)
y
2
ϕ
≥
c
1
y∂
x
ϕ
L
2
(G)
+
c
2
∂
y
ϕ
L
2
(G)
+
c
3
+
2
L
2
(G)
+
(c
4
−
c
3
)
ϕ
2
2
≥
min
{
c
1
,c
2
,c
3
}
ϕ
V
−|
c
4
−
c
3
|
ϕ
L
2
(G)
.
Since by definition
C
0
(G)
is dense in
V
, we may extend the bilinear form
a
κ
(
·
,
·
)
continuously to
V
.
By the abstract well-posedness result Theorem 3.2.2 in the triple of spaces
V
=
V
∗
, we conclude that the weak formulation to the
(transformed) Heston model (
9.20
),
Find
w
∈
L
2
(J
;
V)
∩
H
1
(J
;
L
2
(G))
such that
(∂
t
w,v)
L
2
(G)
=
H
≡
H
∗
⊂
V
⊂
a
κ
(w, v)
f
κ
,v
(9.26)
+
=
V
∗
,V
,
∀
v
∈
V,
a.e. in
J,
w(
0
)
=
0
,
admits a unique solution for every
f
κ
∈
V
∗
.
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